What's the inverse of $x^5 +3x^3 + 2x + 1$?

Question

Let $f$ be a one-to-one function whose inverse is given by: $f^{-1}(x)=x^5+3x^3+2x+1$

1. Compute $f^{−1}(1)$.

My attempt at this yielded a very straightforward answer:

$f^{-1}(1)=1^5+3(1)^3+2(1)+1\\ f^{-1}(1)= 7$

2. Compute $f(1)$.

I have searched around on this Stack Exchanged and discovered that finding the inverse of a degree 5 polynomial is not viable using only algebra level math.

I would greatly appreciate it if someone was able to point me in the right direction. Thanks in advance.

Answer 1

$$f^{-1}(x)=x^5+3x^2+2x+1 \implies f^{-1}(0)=1 \implies f(1)=0.$$ Next $$f^{-1}(1)=7.$$

Answer 2

Without guessing one can see that we are trying to solve:

$$1=x^5+3x^3+2x+1$$

This reduces down to

$$0=x^5+3x^3+2x=x(x^4+3x^2+2)=x(x^2+1)(x^2+2)$$

It's easy to see then that $x=0$ is the only real solution to this, as well as the 4 other solutions if one desired to involve complex numbers.

Answer 3

Inverse here means the inverse function.

$$f^{-1}(x)=x^5+3x^3+2x+1$$

1.)

Your calculation is correct. For $f^{-1}(1)$, we have to set $x=1$ and get $f^{-1}(1)=7$ from the above equation.
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2.)

The preconditions of the task say i.a. that $f$ is a one-to-one function and $f^{-1}$ is its inverse. That means, the inverse of $f^{-1}$ does exist and is $f$.

If $f$ and $f^{-1}$ are inverses of each other, $f(f^{-1}(x))=f^{-1}(f(x))=x$ holds.
If the inverse of a function $f\colon x\mapsto f(x)$ given in closed form does exist, we therefore can try to find the function term of $f^{-1}$ with $y=f^{-1}(x)$ by solving $f(y)=x$ for $y$.

Now $f^{-1}$ is given, and we are looking for its inverse $f$ with $y=f(x)$:

$$f^{-1}(y)=x,$$

$$y^5+3y^3+2y+1=x.$$

The solution $y$ cannot be represented as explicit algebraic function of $x$. But the task needs $f$ only at one point $x=1$.

We set $x=1$ therefore:

$$y^5+3y^3+2y+1=1,$$

$$y^5+3y^3+2y=0,$$

and get $$y_{1,2,3,4,5}={0,-i,i,-i\sqrt{2},i\sqrt{2}}.$$

Because, according to the preconditions, $f$ is a one-to-one function, only one of these values can be $f^{-1}(1)$. To ensure that the task is given correctly, the range of $f^{-1}$ or the domain of $f$ should be specified.